A light bulb and an inductor coil are connected to an ac source through a key as shown in the figure below. The key is closed and after sometime an iron rod is inserted into the interior of the inductor. The glow of the light bulb

In the product

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$=q\overrightarrow{v}\times \left(B\hat{\mathrm{i}}+B\hat{\mathrm{j}}+B_0\hat{\mathrm{k}}\right)$

For $q=1$ and $\overrightarrow{v}=2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+6\hat{\mathrm{k}}$ and $\overrightarrow{F}=4\hat{\mathrm{i}}-20\hat{\mathrm{j}}+12\hat{\mathrm{k}}$

What will be the complete expression for $\overrightarrow{B}$?

A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current-carrying coil in the shape of,

(i) an equilateral triangle of side $a$.
(ii) a square of side $a$.
The magnetic dipole moments of the coil in each case respectively are:

Three resistors having resistances $r_1, r_2$ and ${\mathrm{r}}_3$ are connected as shown in the given circuit. The ratio $\frac{i_3}{i_1}$ of currents in terms of resistances used in the circuit is :